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In the L shape tube shown in the image shouldn't the pressure at all points A, B,C,D be same as atmospheric pressure? I have to find relation between A, B, C and D options

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  • $\begingroup$ Yes, the pressures at A, B, C, and D are atmospheric. So,....? $\endgroup$ Aug 23, 2021 at 12:01

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The "$\omega$" in the drawing suggests that the tube is rotating. In that case there will be a pressure gradient in the horizontal part to provide the centripetal acceleration:

$$ \rho_{\rm air}\omega^2 r = \frac{dP}{dr} $$

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  • $\begingroup$ Shouldn't the water surface in the tank be non-flat, and the depth at D be below the average (rather than above)? $\endgroup$ Aug 23, 2021 at 12:06
  • $\begingroup$ @ChetMiller Maybe only the tube is rotating and not the entire tank. $\endgroup$
    – Tofi
    Aug 23, 2021 at 12:10
  • $\begingroup$ @Tofi That wouldn't produce much of an effect. $\endgroup$ Aug 23, 2021 at 12:17
  • $\begingroup$ @ChetMiller It would still produce a pressure gradient in the tube because the air inside it would be rotating as well. I think, at equilibrium, the pressure at point A would be equal to atmospheric pressure, and the pressures at points B and C are related to that at A by the equation given in the answer. $\endgroup$
    – Tofi
    Aug 23, 2021 at 12:26
  • $\begingroup$ or maybe pressure at point A is related to atmospheric pressure by Bernoulli's equation? idk. $\endgroup$
    – Tofi
    Aug 23, 2021 at 12:29

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